# Getting zero as solution to the 1D wave equation

Your solution $$A_n$$ isn't zero for every $$n$$. Looking at your initial conditions you might expect there to be only one non-zero coefficient since the IC already have the form of a sine wave. When $$n=1$$ you get $$\frac 0 0$$ so you have to be extra careful. Evaluating your solution for $$A_n$$ as a limit: $$\lim_{n\rightarrow 1}\frac{2\sin n\pi}{\pi(n^2-1)}=-1$$ I suspect there is an error in the integral because the answer should be +1. Alternatively you could immediately see from $$\sin\left(\frac \pi l x\right)=\sum_{n=1}^\infty A_n \sin\left(\frac {n\pi} l x\right)$$ that $$A_n=\cases{1&n=1\\0 &n\neq 1}$$

Wolframalpha is likely the source of confusion here: the initial condition is the term with $$n=1$$, i.e. $$A_1 = 1$$, whereas all others are $$A_n=0$$. The equation for the initial condition is one equation with the infinite number of unknowns in the rhs - no surprise that an automatic routine gives an irrelevant answer there.